Find height of a projectile

Question

Height of a Projectile. A stone is thrown directly upward from a height of 30ft with an initial velocity of 60ft/sec. The height of the stone t seconds after it has been thrown is given by the function
s(t)=-16t^2+60t+30. Determine the time at which the stone reaches its maximum height and find the maximum height.

Answer 1

Finding the vertex coordinates (h, k) to get the time and the maximum height,

h = -b/(2a) = 60/(2*16) = 15/8, k = f(h) = 345/4

 

Answer: t = 15/8 sec, and the maximum height is 345/4 ft.

Answer 2

Hey Daniel -- here's a physical reasoning approach you may like ...

 

ave speed from start to peak is 30ft/s ... g of 32ft/s/s takes 1 7/8 sec to slow from

60ft/s to zero at peak ... ht is 30ft plus (ave speed * 1 7/8 sec) ==> 90ft less 30/8 ft

 

==> 86 1/4 ft after 1 7/8 sec ... Best wishes, sir :)

Answer 3

S(t) =  -16 t² + 60t + 30

 

   This is a parabola :  Maximum( -60/ -32 , s( 15/ 8)

 

 

       t   = 15/8

 

 

 

       S( 15/8) = -16(15/8)² +60*(15/8) + 30 = 86.25